An Idea Towards Recovering Time

[Aethelwulf]

The Wheeler de Witt equation (as you can see from two previous essays on the subject)

 

http://scienceforums.com/topic/26897-true-time-evolution-in-gr-and-the-wheeler-de-witt-equation/

 

http://scienceforums.com/topic/27016-complexification-and-the-wave-function-of-the-universe/

 

predicts that the time derivative describing the universal wave function vanishes. This is just not acceptable because time is part of the inherent complexification of the theory, but the complex nature seems to vanish. How can we get time back?

 

Time would appear as the motion of particles and their respective, or better said their relative motions. Time appears then as arising from changing dynamical systems.

 

Consider a simple spacetime interval as:

 

[math]d\tau^2 = dt^2 - d\vec{x}^2[/math]

 

Where we have set [math]c=1[/math] in this case. You actually calculate the length of a worldline by taking into consideration the integral

 

[math]L(W) = \int_W d\tau[/math]

 

You can, it was shown to me a while ago now, that worldines can be written in terms of time by the chain rule. Doing so, you can rewrite the time derivatives as dots on your variables and can end up with

 

[math]L(W) = \int_{t_0}^{t_1} \sqrt{1 - ||\dot{x}||^2}\ dt[/math]

 

From here, you would calculate the Langrangian by simply multiplying mass into the equation, so we would have

 

[math]\mathcal{L} = -M\sqrt{1 - ||\dot{x}||^2}[/math]

 

Now in my equation, we have been using the generalized velocity, and can be freely exchanged now to make the above equation into

 

[math]\mathcal{L}(\dot{q}\dot{q}) = -M\sqrt{1 - \dot{q}\dot{q}}[/math]

 

Now, the canonical momentum part in my equation can be written as

 

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}} = \frac{M\dot{q}}{\sqrt{1 - \dot{q}\dot{q}}}[/math]

 

This is relativistic and is incorporated as one can see, into the idea of worldlines. Now, in my equation, I decided to multiply the momentum with distance. Of course, this was just the quantum action [math]\hbar[/math], but ignoring that fact for now, we wish to calculate the distance really as a displacement of all the particles in the universe [math]d_i[/math] using Barbour's approach. Doing so, will require an integral.

 

Taking the integral of the equation, which ''cuts up'' or ''slices'' a worldline for a particle, then the distance will be small [math]\delta d[/math] for a particle which is the way alluded to perhaps by Barbour when calculating displacements of particles - position is relative in quantum mechanics. Time may appear then as the displacements of particles over ''moments.''

 

Remembering that

 

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}} = \frac{M\dot{q}}{\sqrt{1 - \dot{q}\dot{q}}}[/math]

 

We can multiply the momentum part with a displacement symbol [math]\delta d_i[/math] where the i-th component represents your i-th particle, then multiply this with the nabla operator squared (laplace operator) being hit on a wave function gives

 

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}}(\delta d_i) \nabla^2 \psi = \dot{m}\psi[/math]

 

This is a mass-flow equation. To get some idea of how this idea works, Barbour brought his idea's on time being the displacements of particles in this wonderful video

 

Edited December 3, 2012 by Aethelwulf

[Aethelwulf]

I meant to say, complexification may arise as

 

[math]-i \hbar \nabla^2 \psi = \dot{m}\psi[/math]

 

Where [math]\hbar[/math] plays the quantum role of

 

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}}(\delta d_i)[/math]

 

Complexifying this term is not too different to the momentum operator which exists as

 

[math]-i \hbar \nabla[/math]

 

except for the operator which is taken to the second power. Maybe there is another way to do it which I need to figure out.

Edited December 3, 2012 by Aethelwulf

[Aethelwulf]

How do you derive

 

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}}(\delta d_i) \nabla^2 \psi[/math]

 

Concerning the part [math]\nabla^2 \psi[/math]

 

?

 

 

Let us first concern ourselves with the nabla operator

 

[math]\nabla = \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}[/math]

 

Hitting this with a wave function yields

 

[math]\nabla \psi = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} + \frac{\partial \psi}{\partial z}[/math]

 

Hit the nabla operator with

 

[math]\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}[/math]

 

Gives

 

[math]\nabla(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}) \psi = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2}[/math]

 

this is the same as

 

[math]\hat{e}_x \frac{\partial^2 \psi}{\partial x^2} + \hat{e}_y \frac{\partial^2 \psi}{\partial y^2} + \hat{e}_z \frac{\partial^2 \psi}{\partial z^2}[/math]

 

Which reduces to simply

 

[math](\nabla \cdot \nabla)\psi[/math]

 

Or

 

[math]\nabla^2 \psi[/math]

 

The term alone [math]\nabla \psi[/math] is actually the same as saying

 

[math]ik_x \psi e_x + ik_y \psi e_y + ik_z \psi e_z = \frac{i}{\hbar} \hat{P}\psi[/math]

 

thus is the same as saying [math]\nabla^2 \psi[/math] when you gather all the terms together

 

[math]\frac{i}{\hbar} \hat{P}(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z})\psi[/math]

Edited December 3, 2012 by Aethelwulf

[Aethelwulf]

This justifies why there should be an imaginary number attached to the equation

 

[math]-i\frac{\partial \mathcal{L}}{\partial \dot{q}} \delta d_i \nabla^2 \psi = \dot{m}\psi[/math]

 

where the [math]\dot{m}[/math] is obviously taken with it's time derivative.

Edited December 3, 2012 by Aethelwulf

[Aethelwulf]

''

The term alone [math]\nabla \psi[/math] is actually the same as saying

 

[math]ik_x \psi e_x + ik_y \psi e_y + ik_z \psi e_z = \frac{i}{\hbar} \hat{P}\psi[/math]

''

 

I will explain this better. The plane wave solution to the Schrodinger equation is

 

[math]\Psi = e^{i(k \cdot r - \omega t)}[/math]

 

The gradient of this is

 

[math]\hat{e}_x \frac{\partial \psi}{\partial x} + \hat{e}_y \frac{\partial \psi}{\partial y} + \hat{e}_z \frac{\partial \psi}{\partial z} = ik_x \psi e_x + ik_y \psi e_y + ik_z \psi e_z[/math]

 

[math]= \frac{i}{\hbar}(p_x e_x + p_y e_y + p_z e_z)\psi = \frac{i}{\hbar} \hat{P}\psi[/math]

 

and because [math]e_x, e_y[/math] and [math]e_z[/math] are the base space, we can see that

 

[math]\hat{P} = -i \hbar \nabla[/math]

 

Which makes our momentum in position space.

Edited December 3, 2012 by Aethelwulf

[Aethelwulf]

Keeping in mind that

 

[math]\frac{\partial}{\partial x}(\frac{\partial}{\partial x}) = \frac{\partial^2}{\partial x^2}[/math]

 

Thus

 

[math]-i \hbar \nabla(\frac{\partial}{\partial y}+ \frac{\partial}{\partial x} + \frac{\partial}{\partial z})[/math]

 

Is the same as

 

[math]-i \frac{\partial \mathcal{L}}{\partial \dot{q}} (\delta d_i) \nabla^2[/math]

Edited December 3, 2012 by Aethelwulf

[Aethelwulf]

When the Universe was created, it truly was timeless. It was flooded with radiation and according to relativity, radiation do not act as clocks. Before spacetime properly expanded from a point, it began then timeless as well since space did not exist. A universe born without time is such a beautiful thing to consider, but time appeared when matter began to emerge from the decay of this radiation-flooded universe. Time seems to be an emergent property - in fact spacetime geometry is inherently emergent if we take geometrogenesis seriously. In this forgotten post of mine, I have been able to demonstrate an equation which not only takes into consideration re-complexifying the mass-flow equation I presented in this thread.

 

Of course, I have noted plenty of times, that if one actually quantizes the Einstein Field equations, you get back what is called the Wheeler de Witt equation which is devoid of complexification (the presence of an imaginary number coefficient which would normally be found in a Schrodinger like equation describing the energy of a system). Some have considered this to mean that gravity itself might be inherently real unlike all the other forces. But an interesting thing to keep in mind, is that if geometry is emergent, then so is gravity and therefore gravity is not fundamental which must imply that the Wheeler de Witt equation is describing an unreal situation - it would mean that you cannot fundamentally-describe General Relativity in nature.

[Aethelwulf]

It can also be argued that my equation

 

 

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}}(\delta d_i) \nabla^2 \psi = \dot{m}\psi[/math]

 

Might be more correctly given as the four dimensional case

 

 

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}}(\delta d_i) \Box \psi = \dot{m}\psi[/math]

 

Where the [math]\Box[/math] is the d'Alembertian. Sometimes it is more aesthetically pleasing on the eye to write the d'Alembertain as

 

 

[math]\frac{\partial \mathcal{L}}{\partial \dot{q}}(\delta d_i) \nabla^{\mu} \nabla_{\mu} \psi = \dot{m}\psi[/math]

Edited March 27, 2013 by Aethelwulf